Let $S$ be the region between two concentric spheres of radii $4$ and $6$, both centered at the origin. What is the triple integral of $f(\rho) = \rho^2$ over $S$ in spherical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_4^6 \rho^4\sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $ \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^3 \sin^2(\varphi)\cos(\theta) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_4^6 \rho^3 \sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_0^6 \rho^3 \sin^2(\varphi)\sin(\theta) \, d\rho \, d\theta \, d\varphi$
Explanation: The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the concentric spheres with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. Here, the region $S$ ranges from a radius of $4$ to a radius of $6$. Therefore, we want $4 < \rho < 6$. $ \int_0^\pi \int_0^{2\pi} \int_4^6 \cdots \, d\rho \, d\theta \, d\varphi$ The scalar field we want to integrate over depends on the distance from the origin $\rho$, so we don't need to make any substitutions with spherical coordinates. Our integral looks like this now: $ \int_0^\pi \int_0^{2\pi} \int_0^1 \rho^2 \cdots \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_0^1 \rho^4 \sin(\varphi) \, d\rho \, d\theta \, d\varphi$